Lemma 65.16.3. Let $\mathit{Sch}_{fppf}$ be a big fppf site. Let $S \to S'$ be a morphism of this site. The construction above give an equivalence of categories
\[ \left\{ \begin{matrix} \text{category of algebraic}
\\ \text{spaces over }S
\end{matrix} \right\} \leftrightarrow \left\{ \begin{matrix} \text{category of pairs }(F', F' \to S)\text{ consisting}
\\ \text{of an algebraic space }F'\text{ over }S'\text{ and a}
\\ \text{morphism }F' \to S\text{ of algebraic spaces over }S'
\end{matrix} \right\} \]
Proof.
Let $F$ be an algebraic space over $S$. The functor from left to right assigns the pair $(j_!F, j_!F \to S)$ to $F$ which is an object of the right hand side by Lemma 65.16.1. Since this defines an equivalence of categories of sheaves by Sites, Lemma 7.25.4 to finish the proof it suffices to show: if $F$ is a sheaf and $j_!F$ is an algebraic space, then $F$ is an algebraic space. To do this, write $j_!F = U'/R'$ as in Lemma 65.9.1 with $U', R' \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S')_{fppf})$. Then the compositions $U' \to j_!F \to S$ and $R' \to j_!F \to S$ are morphisms of schemes over $S'$. Denote $U, R$ the corresponding objects of $(\mathit{Sch}/S)_{fppf}$. The two morphisms $R' \to U'$ are morphisms over $S$ and hence correspond to morphisms $R \to U$. Since these are simply the same morphisms (but viewed over $S$) we see that we get an étale equivalence relation over $S$. As $j_!$ defines an equivalence of categories of sheaves (see reference above) we see that $F = U/R$ and by Theorem 65.10.5 we see that $F$ is an algebraic space.
$\square$
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